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Kruka [31]
3 years ago
12

WWhat do these substances have in common?

Chemistry
2 answers:
olga2289 [7]3 years ago
8 0

Answer; Option A is correct

Explanation:

  • Matter is substance which is made up of atoms
  • Matter is a substance which occupies physical space

All the given four substance are matter :

  • Apple juice is a mixture.
  • Helium gas is a pure substance.
  • Salt water is mixture.
  • Copper wire is a pure substance.

But all of them are composed of atoms.


Mariana [72]3 years ago
4 0

The answer is that they are Pure substances

You might be interested in
Example of electrovalent bond​
olchik [2.2K]

Explanation:

<h3>MgCl2, CaCl2, MgO, Na2S, CaH2, AlF3, NaH, KH, K2O, KI, RbCl, NaBr, CaH2 etc.</h3>

<h2>hope it helps.</h2><h2>stay safe healthy and happy..</h2>
4 0
3 years ago
Read 2 more answers
If a 95.27 mL sample of acetic acid (HC2H3O2) is titrated to the equivalence point with 79.06 mL of 0.113 M KOH, what is the pH
KATRIN_1 [288]

Answer:

8.73

Explanation:

The concentration of acetic acid can be determined as follows:

M_1V_1 = M_2V_2\\(KOH) = (CH_3COOH)

M_{KOH}=0.113 M\\V_{KOH}=79.06 mL

V_{CH_3COOH}=95.27 \\\\M_{CH_3COOH)=?????

M_{CH3COOH} = \frac{M_{KOH}*V_{KOH}}{V_{CH_3COOH}}

M_{CH3COOH} = \frac{0.113*79.06}{95.27}

M_{CH3COOH} = 0.094 M

Moles of CH_3COOH = 95.27* 10^{-3}* 0.094

=0.0090 moles

Moles of  KOH = 79.06*10^{-3}*0.113

= 0.0090 moles

The equation for the reaction can be expressed as :

CH_3COOH     +      KOH     ----->      CH_3COO^{-}K^+      +     H_2O

Concentration of CH_3COO^{- ion = \frac{0.0090}{Total volume (L)}

= \frac{0.0090}{(95.27+79.06)} *1000

= 0.052 M

Hydrolysis of  CH_3COO^{- ion:

CH_3COO^{-      +       H_2O      ----->      CH_3COOH       +     OH^-

K = \frac{K_w}{K_a} = \frac{x*x}{0.052-x}

⇒    \frac{10^{-14}}{1.82*10^{-5}}= \frac{x*x}{0.052-x}

=     0.5494*10^{-9}= \frac{x*x}{0.052-x}

As K is so less, then x appears to be a very infinitesimal small number

0.052-x ≅ x

0.5494*10^{-9}= \frac{x^2}{0.052}

x^2 = 0.5494*10^{-9}*0.052

x^2 = 0.286*10^{-10

x = \sqrt{0.286*10^{-10

x =0.535*10^{-5}M

[OH] = x =0.535*10^{-5}

pOH = -log[OH^-]

pOH = -log[0.535*10^{-5}]

pOH = 5.27

pH = 14 - pOH

pH = 14 - 5.27

pH = 8.73

Hence, the pH of the titration mixture = 8.73

8 0
3 years ago
Which of the following values is measured the LEAST precisely? * 22mL 45.2mL 50mL 100mLchem​
nekit [7.7K]

Answer:

If you're looking at the data as a whole, it would most likely be 100ml.

Explanation: The definition of precise is data close together so 100ml is furthest away from the other recorded numbers

7 0
3 years ago
A balloon filled with helium has a volume of 6.9 L. What is the mass, in grams, of helium in the balloon?
balandron [24]

Answer : The mass of helium gas in the balloon is, 1.23 grams.

Explanation : Given,

Volume of helium gas = 6.9 L

First we have to calculate moles of helium gas at STP.

As we know that, 1 mole of substance occupy 22.4 L volume of gas.

As, 22.4 L volume of helium gas present in 1 mole of helium

So, 6.9 L volume of helium gas present in \frac{6.9L}{22.4L}\times 1mole=0.308mole of helium

Now we have to calculate the mass of helium gas.

\text{Mass of He gas}=\text{Moles of He gas}\times \text{Molar mass of He gas}

Molar mass of He gas = 4 g/mol

\text{Mass of He gas}=0.308mol\times 4g/mol=1.23g

Thus, the mass of helium gas in the balloon is, 1.23 grams.

6 0
3 years ago
1. Determine the final volume of a 1.5 M HCl solution prepared from 20.0 mL
nadezda [96]

Explanation:

Because molarity is classified as moles of solute per liter of water, dilution of the water may result in a reduction of its concentration.

Therefore, because the amount of moles of solute has to be constant for dilution, you will use the molarity and volume of that same target solution to calculate how many moles of solute will be present in the sample of the stock solution that you dilute.

c  =   \frac{n}{v}

⇒  n = c ⋅  V

n_{HCL} =  0.250 M  ⋅  6.00 L  = 1.5 moles HCl

Now all you have to do is figure out what volume of   6.0 M  stock solution will contain  1.5  moles of hydrochloric acid

c = \frac{n}{v}

V = \frac{n}{c}

V_{Stock} = \frac{1.5 moles}{6.0 \frac{moles}{L} }   = 0.25 L

Expressed in milliliters, the answer will be

V_{Stock} = 250ML →  rounded to two sig figs

7 0
3 years ago
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